Homework Problems

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Homework Problems Chapter 4 Homework Problems
2, 8, 11, 14, 20, 22, 25, 28, 29, 30, 32, 42, 44,
48, 52, 54, 64, 68, 82, 84, 94, 124, 132, 150
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CHAPTER 4 Reactions in Aqueous Solution
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Solutions A solution is a homogeneous mixture of
two or more pure chemical compounds. Solutions
may be solids, liquids, or gases. Solvent - The
major component of a solution (if there is
one) Solute(s) - The minor component(s) of a
solution Example A solution is prepared by
adding a small amount of sodium chloride (NaCl)
to water. solvent water (major
component) solute sodium chloride Aqueous
solution - Solution where water is the solvent.
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Electrolytes and Nonelectrolytes Many chemical
substances dissolve in water. A substance that
forms ions when dissolved in water is called an
electrolyte, while a substance that does not form
ions when dissolved in water is called a
nonelectrolyte. Electrolytes can be divided into
two categories Strong electrolyte - Completely
dissociates (falls apart) when added to solution,
to form ions. Weak electrolyte - Partially
dissociates (falls apart) when added to solution,
to form ions.
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Classification of Compounds as Electrolytes (in
Water) We previously divided compounds into two
general categories Ionic compounds -
Collections of cations (positive ions) and anions
(negative ions) in a crystal structure. Usually
form from a metal (or cation group) and a
nonmetal (or anion group). Molecular compounds -
Exist as individual molecules. Usually form from
two or more nonmetals. We may use this as a
starting point in classifying compounds as strong
electrolytes, weak electrolytes, and
nonelectrolytes.
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Ionic Compounds Ionic compounds - All ionic
compounds are strong electrolytes. Examples NaCl
(s) ? Na(aq) Cl-(aq) KOH(s) ? K(aq)
OH-(aq) Ca(NO3)2(s) ? Ca2(aq) 2
NO3-(aq) Note that an insoluble ionic compound
is considered a strong electrolyte. This is
because even though very little of the compound
dissolves in water, whatever does dissolve
completely ionizes.
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Molecular Compounds Strong acids - Molecular
compounds that are strong acids are also strong
electrolytes. Examples HCl(aq) ? H(aq)
Cl-(aq) HNO3(aq) ? H(aq) NO3-(aq) Weak
acids and bases - Molecular compounds that are
weak acids or weak bases are also weak
electrolytes. Note that an acid forms H ion
when added to water, and a base forms OH-
ion. Examples HF(aq) ? H(aq) F-(aq) NH3(aq)
H2O(l) ? NH4(aq) OH-(aq) The double arrow
? indicates a reaction that proceeds in both
the for-ward and reverse direction, establishing
equilibrium.
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Molecular Compounds (Continued) Other molecular
compounds - Molecular compounds that are neither
acids nor bases are nonelectrolytes. Examples C6
H12O6(s) ? C6H12O6(aq) CH3OH(l) ?
CH3OH(aq) Note that water (H2O) is classified as
a nonelectrolyte, even though a small number of
H and OH- ions are formed in pure water.
H2O(l) ? H(aq) OH-(aq) For pure water at T
25 there are only 2 x 10-7 M of ions present.
This is small enough that water is classified as
a nonelectrolyte.
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Flow Chart For Classifying Compounds
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Precipitation Reactons A precipitation
reactions is a reaction that forms a solid (a
precipitate) out of soluble reactants. Examples
Pb(NO3)2(aq) 2 KCl(aq) ? PbCl2(s) 2
KNO3(aq) 3 NaOH(aq) FeBr3(aq) ? Fe(OH)3(s)
3 NaBr(aq)
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Solubility (Water) Solid compounds are divided
into two general categories. Soluble compound -
a compound that will dissolve in
water Insoluble compound - a compound that will
not dissolve in water The above classification
scheme is an oversimplification. Most
insoluble compounds will dissolve to a slight
extent. For soluble compounds there will be a
limit as to the amount of compound that will
dissolve in a given amount of water. We can
also discuss solubility in other solvents.
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Why Ionic Compounds Dissolve in Water The
forces holding an ionic compound together are the
strong electrical attraction that exists between
cations and anions. It is therefore somewhat
surprising that ionic compounds will dissolve in
water. The reason some ionic compounds will
dissolve in water is because the water molecules
have a partial negative charge on the oxygen atom
(?-) and partial positive charges on the hydrogen
atoms (?), where ? indicates a small positive
or negative charge. The reason these partial
charges exist will be discussed later in the
semester. Because cations and anions can
strongly interact with the negative and positive
partial charges on several water molecules at the
same time, it is possible for the ions in an
ionic compound to be pulled out of the solid and
go into solution.
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Hydration and Solvation The process of
surrounding a cation or anion with water
molecules to stabilize them in aqueous solution
is called hydration. Often these ions will be
surrounded by a definite number of water
molecules, called a hydration shell. The more
general process of surrounding ions or molecules
with solvent molecules in a solution is called
solvation. Water is unusual because the
interaction of ions with water molecules is
much stronger than the interactions that occur
in most other solvents.
hydration of
anion
hydration of
cation
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Solubility Rules a) All compounds with group IA
cations, or NH4 (ammonium ion), are soluble. b)
All compounds containing NO3- (nitrate ion),
ClO4- (perchlorate ion), ClO3- (chlorate ion), or
C2H3O2- (acetate ion) are soluble. c) Most
compounds containing Cl-, Br-, and I- are
soluble. (EXCEPTIONS Cations Ag, Pb2,
Hg22). d) Most compounds containing SO42-
(sulfate ion) are soluble. (EXCEPTIONS Cations
Ag, Pb2, Hg22, Ca2, Sr2, Ba2). e) Most
compounds containing OH- (hydroxide ion) are
insoluble. (EXCEPTIONS Cations from group 1A,
Ca2, Sr2, Ba2). f) Most other compounds are
insoluble. (EXCEPTIONS Compounds covered by
rules a-e).
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Use of Solubility Rules KClO4 (NH4)2SO4
CuCl2 PbI2 NiSO4 Mg(OH)2 Zn(IO3)2

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Use of Solubility Rules KClO4 soluble Rules
a and b (NH4)2SO4 soluble Rule
a CuCl2 soluble Rule c PbI2 insoluble Rule
c NiSO4 soluble Rule d Mg(OH)2 insoluble Ru
le e Zn(IO3)2 insoluble Rule f
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Determining Whether a Precipitation Reaction Will
Occur When solutions of two soluble substances
mix we will not always form a precipitate. We
may use the solubility rules to decide if a
precipitate will form. 1) Write the ions formed
from the starting compounds. 2) Look at all
possible combinations of cations and anions.
a) If there is a combination of ions that forms
an insoluble compound, a precipitate will form
and a precipitation reaction occurs. b) If
there is not a combination of ions that forms an
insoluble compound, then there will be no
precipitation reaction.
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Examples For each of the following mixtures of
two solutions will a precipitate form? Note that
all of the solutes are themselves
soluble. Solution A solution
B precipitate(?) 1) NaCl(aq) AgNO3(aq) 2) NaCl
(aq) CuBr2(aq)
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Solution A solution B precipitate(?) 1) NaCl(a
q) AgNO3(aq) yes, AgCl(s) Na Cl- Ag NO3-
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
2) NaCl(aq) CuBr2(aq) no Na Cl- Cu2 Br-
No reaction.
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Net Ionic Equations To this point we have
written chemical equations in terms of neutral
species (molecules, or formula units). However,
in aqueous solution it is often more appropriate
to write equations in terms of ions. There are
three ways we normally write chemical
equations 1) Molecular equation. All reactants
and products are written in terms of their
chemical formula. 2) Total ionic equation.
Strong electrolytes (strong acids, strong soluble
bases, and soluble ionic compounds) are written
as ions. 3) Net ionic equation. The total ionic
equation after removing spectator ions. A
spectator ion is an ion that appears on both the
reactant and product side of the equation, and
which does not participate directly in the
chemical reaction.
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Net Ionic Equation (Example) We may show the
various ways of writing chemical reactions as
follows Molecular AgNO3(aq) NaCl(aq) ?
NaNO3(aq) AgCl(s) Total Ionic Ag(aq)
NO3-(aq) Na(aq) Cl-(aq) ? Na(aq)
NO3-(aq) AgCl(s) Net Ionic Ag(aq) Cl-(aq)
? AgCl(s)
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Finding the Total and New Ionic Equations We
find the total and net ionic equations as
follows. 1) Begin with a balanced molecular
equation. 2) Break apart strong electrolytes
into ions to obtain the total ionic equation. 3)
Cancel ions that appear on both sides of the
equation to obtain the net ionic
equation. Example molecular Pb(NO3)2(aq) 2
KI(aq) ? PbI2(s) 2 KNO3(aq)
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molecular Pb(NO3)2(aq) 2 KI(aq) ? PbI2(s) 2
KNO3(aq) total ionic Pb2(aq) 2 NO3-(aq) 2
K(aq) 2 I-(aq) ? PbI2(s) 2 K(aq) 2
NO3-(aq) net ionic Pb2(aq) 2 I-(aq) ? PbI2(s)
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By writing the net ionic equation we are able to
focus on the important parts of a chemical
reaction and observe similarities in
reactions. mol. AgNO3(aq) NaCl(aq) ? AgCl(s)
NaNO3(aq) total Ag(aq) NO3-(aq) Na(aq)
Cl-(aq) ? AgCl(s) Na(aq)
NO3-(aq) net Ag(aq) Cl-(aq) ?
AgCl(s) mol. AgNO3(aq) KCl(aq) ? AgCl(s)
KNO3(aq) total Ag(aq) NO3-(aq) K(aq)
Cl-(aq) ? AgCl(s) K(aq)
NO3-(aq) net Ag(aq) Cl-(aq) ? AgCl(s)
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Net Ionic Equation When No Precipitate
Forms Writing the net ionic equation makes it
clear when precipitation reactions do or do not
occur. mol. 2 KBr(aq) Zn(NO3)2(aq) ? 2
KNO3(aq) ZnBr2(aq) K Br- Zn2 NO3-
total 2 K(aq) 2 Br-(aq) Zn2(aq) 2
NO3-(aq) ? 2 K(aq) 2 NO3-(aq) Zn2(aq)
2 Br-(aq) All the ions are spectator ions, and
so no reaction occurs.
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Acids and Bases (Arrhenius) Arrhenius
definition acid - A substance which, when added
to water, forms H ion base - A substance which,
when added to water, forms OH- ion Examples HCl(
aq) ? H(aq) Cl-(aq) HF(aq) ? H(aq)
F-(aq) NaOH(s) ? Na(aq) OH-(aq) NH3(aq)
H2O(l) ? NH4(aq) OH-(aq) The advantage of
the Arrhenius definition is that it is simple and
easy to implement. The disadvantage is that it
is tied in to a particular solvent (water) and is
not a general definition.
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Strong and Weak Acids An acid is a substance
that forms H ion when added to water (Arrhenius
definition). A strong acid forms one H ion for
each acid molecule. A weak acid forms much less
than one H ion for each acid molecule. Examples
HClO4(aq) ? H(aq) ClO4-(aq) strong HNO2(aq)
? H(aq) NO2-(aq) weak There are 7 common
strong acids binary ternary
HCl HI HClO3 HNO3
HBr HClO4 H2SO4 (1st proton) All other
acids can be assumed to be weak acids.
H2SO4(aq) ? H(aq) HSO4-(aq)
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Strong and Weak Bases A base is a substance that
forms OH- ion when added to water (Arrhenius
definition). A strong soluble base is a soluble
hydroxide compound that completely dissociates
when added to water. An insoluble base is an
insoluble hydroxide compound. There are also a
few substances that act as weak bases in
solution. These substances form much less than
one OH- ion per base molecule. Examples KOH(s) ?
K(aq) OH-(aq) strong soluble
base Fe(OH)3(s) ? no reaction insoluble
base NH3(aq) H2O(l) ? NH4(aq) OH-(aq) weak
base There are 8 common strong soluble bases
Group 1A - LiOH, NaOH, KOH, RbOH, CsOH
Group 2A - Ca(OH)2, Sr(OH)2, Ba(OH)2 All other
hydroxide compounds are insoluble bases.
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Acids and Bases (Bronsted) Bronsted
definition acid - a proton (H) donor base - a
proton (H) acceptor
HF(aq) H2O(l) ? H3O(aq) F-(aq)
acid base
In the Bronsted theory, in an acid-base
reaction an acid donates a proton, while a base
accepts a proton. In addition, in Bronsted
theory acids form hydronium ion (H3O ion)
instead of hydrogen ion (H ion) when added to
water. The Bronsted definition of an acid and a
base is more general than the Arrhenius
definition.
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Examples of Acids (Bronsted) Strong acid
HCl(aq) H2O(l) ? H3O(aq)
Cl-(aq) acid HCl base H2O Weak acid
HC2H3O2(aq) H2O(l) ? H3O(aq)
C2H3O2-(aq) acid HC2H3O2 base H2O The
second reaction goes in both directions.
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Examples of Bases (Bronsted) Strong base
NaOH(s) ? Na(aq) OH-(aq) Considered an
ionization reaction in Bronsted theory, not an
acid-base reaction. Weak base NH3(aq) H2O(l)
? NH4(aq) OH-(aq) base NH3 acid H2O T
he second reaction goes in both directions.
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Example Chlorous acid (HClO2) is a weak acid.
Pyridine (C5H5N) is a weak base. Indicate the
behavior of these two substances when added to
water, according to Bronsted theory.
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Example Chlorous acid (HClO2) is a weak acid.
Pyridine (C5H5N) is a weak base. Indicate the
behavior of these two substances when added to
water, according to Bronsted theory. HClO2(aq)
H2O(l) ? H3O(aq) ClO2-(aq) C5H5N(aq)
H2O(l) ? C5H5NH(aq) OH-(aq) Notice that,
depending on the reaction, water can act as
either a Bronsted base or a Bronsted acid. A
substance that can act as either a Bronsted acid
or a Bronsted base is called amphoteric.
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Polyprotic Acid A polyprotic acid has two or
more ionizable protons that can be donated in an
acid-base reaction. Monoprotic (one ionizable
proton) HCl, HNO2, HC2H3O2 Diprotic (two
ionizable protons) H2CO3, H2SO4 Triprotic
(three ionizable protons) H3PO3 Example H2SO4
(sulfuric acid) 1st proton H2SO4(aq)
H2O(l) ? H3O(aq) HSO4-(aq) 2nd
proton HSO4-(aq) H2O(l) ? H3O(aq)
SO42-(aq) Example H2CO3 (carbonic acid) 1st
proton H2CO3(aq) H2O(l) ? H3O(aq)
HCO3-(aq) 2nd proton HCO3-(aq) H2O(l) ?
H3O(aq) CO32-(aq)
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Other Amphoteric Species We previously pointed
out that depending on the reaction taking place
water can act as either a Bronsted acid or a
Bronsted base. Other species have this same
property. Example HCO3- ion as
acid HCO3-(aq) OH-(aq) ? CO32- (aq) H2O(l)
as base HCO3-(aq) H3O(aq) ? H2CO3(aq)
H2O(l) Like water, HCO3- ion is classified as
amphoteric. Note that the above property of the
hydrogen carbonate ion (and other amphoteric
species) can be useful in acid-base systems.
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Dibasic Base A dibasic base is a strong base
that releases two OH- ions per formula unit of
base. Example Ba(OH)2 (barium
hydroxide) Ba(OH)2(s) ? Ba2(aq) 2 OH-(aq)
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Acid-Base (Neutralization) Reactions An
acid-base reactions (or neutralization reaction)
is the reaction of an acid with a base. These
reactions usually form a salt and water. salt -
An ionic compound that can be formed by the
reaction of an acid with a base. Examples HNO2(a
q) NaOH(aq) ? NaNO2(aq) H2O(l) 2 HClO4(aq)
Ca(OH)2(aq) ? Ca(ClO4)2(aq) 2 H2O(l) 2
HBr(aq) Cu(OH)2(s) ? CuBr2(aq) 2 H2O(l)
The last reaction above illustrates why
insoluble hydroxide compounds are considered
insoluble bases. Note that ionic compounds other
than hydroxides (OH-) or oxides (O2-) are
classified as salts, as they can be formed from
acid-base reactions.
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Strong Acid Strong Base All strong acid
strong base reactions have the same net ionic
equation. mol. 2 HClO4(aq) Sr(OH)2(aq) ?
Sr(ClO4)2(aq) 2 H2O(?) total 2 H(aq) 2
ClO4-(aq) Sr2(aq) 2 OH-(aq) ? Sr2(aq)
2 ClO4-(aq) 2 H2O(?) net H(aq) OH-(aq) ?
H2O(?) mol. HCl(aq) NaOH(aq) ? NaCl(aq)
H2O(?) total H(aq) Cl-(aq) Na(aq) OH-(aq)
? Na(aq) Cl-(aq) H2O(?) net H(aq)
OH-(aq) ? H2O(?)
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Net Ionic Equation for Weak Acid or Base
Reaction When the acid or base in an acid base
reaction is weak, the net ionic equation will be
slightly more complicated. mol. HNO2(aq)
KOH(aq) ? KNO2(aq) H2O(?) mol. HBr(aq)
NH3(aq) ? NH4Br(aq)
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mol. HNO2(aq) KOH(aq) ? KNO2(aq)
H2O(?) total HNO2(aq) K(aq) OH-(aq) ?
K(aq) NO2-(aq) H2O(?) net HNO2(aq) OH-(aq)
? NO2-(aq) H2O(?) mol. HBr(aq) NH3(aq) ?
NH4Br(aq) total H(aq) Br-(aq) NH3(aq) ?
NH4(aq) Br-(aq) net H(aq) NH3(aq) ?
NH4(aq) Notice that we do not break apart the
weak acid or weak base, as they are weak
electrolytes, and therefore exist mainly as
molecules rather than ions.
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Oxidation Number An oxidation number is a number
assigned to an atom in a molecule or ion
indicating whether that atom is electron rich
(negative oxidation number), electron poor
(positive oxidation number), or neutral
(oxidation number of zero). Oxidation numbers do
not indicate charges of atoms, but changes in
oxidation number indicate whether atoms have
gained or lost electron density. A reaction
where some of the oxidation numbers change in
going from reactants to products is called an
oxidation-reduction (redox) reaction.
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Rules For Assigning Oxidation Numbers 1) The sum
of the oxidation numbers of a species is equal to
the charge of the species. 2) Atoms in elemental
forms have an oxidation number of 0. 3) Hydrogen
has an oxidation number of 1 when bonded with
nonmetals (molecular compounds) and -1 when
combined with metals (ionic compounds). 4)
Oxygen usually has an oxidation number of -2.
Exceptions Elemental forms, peroxides (H2O2),
OF2. H2O2 H 1, O -1 OF2 F -1, O 2 5)
Halogens. F in compounds is always -1. Other
halogens are often -1, but in compounds with
oxygen or other halogens the oxidation number may
take on different values. 6) For assigning
oxidation numbers in ionic compounds, it is
useful to break the compounds up into ions.
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Examples Assign oxidation numbers for each
atom in the following substances. H2O Fe(NO3)2
SO42- O3 HClO2 MnO4-
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H2O H 1 O -2 Fe(NO3)2 Fe2 and
NO3- so Fe 2, O -2, and so N 5 SO42- O
-2, and so S 6 O3 elemental form, so O
0 HClO2 H and ClO2- (even though
molecular) so H 1, O -2, and so Cl
3 MnO4- O -2, and so Mn 7
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Common Oxidation Numbers
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Oxidation-Reduction Reactions In an
oxidation-reduction (redox) reaction the
oxidation numbers of some of the atoms involved
in the reaction change when going from reactants
to products. 2 Fe2O3(s) 3 C(s) ? 4 Fe(s) 3
CO2(g) 3 -2
0 0 4
-2 We use the following terms in reference to
redox reactions. Oxidation - An increase in the
value for the oxidation number. Reduction - A
decrease in the value for the oxidation
number. There will always be one oxidation and
one reduction process per reaction. C 0 to 4,
and so is oxidized. Fe 3 to 0, and so is
reduced.
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Oxidizing and Reducing Agents We call the
species that is oxidized in a redox reaction the
reducing agent, and the species that is reduced
in a redox reaction an oxidizing agent. This
seems confusing, but is based on the idea that
there must be both an oxidation and a reduction
taking place in a redox reaction. C(s)
O2(g) ? CO2(g) 0 0
4 -2 In this reaction carbon is
oxidized (from 0 to 4) and so C is a reducing
agent, and oxygen is reduced (from 0 to -2) and
so O2 is an oxidizing agent. SO2(g) 2
Fe3(aq) 2 H2O(l) ? 2 Fe2(aq) SO42-(aq) 4
H(aq) 4 -2 3
1 -2 2
6 -2
1 In this reaction S is oxidized (from 4 to
6) and so SO2 is a reducing agent, and Fe is
reduced (from 3 to 2) and so Fe3 is an
oxidizing agent.
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Activity Series The higher the element in the
activity series, the easier it is to oxidize.
The lower the anion in the activity series, the
easier it is to reduce.
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Active and Noble Metals Some metals in the
activity series (those at or near the top of the
series) are so reactive that they are never found
in the metallic state in nature. Such metals are
called active metals. The alkali metals and
alkaline earth metals are active metals.
Metals at or near the bottom of the activity
series are not very reactive, and so are often
found in the metallic state in nature. Such
metals are called noble metals. Examples are
gold (Au), platinum (Pt), and mercury (Hg).
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Use of the Activity Series The activity series
can be used to predict which metals will reduce a
particular ion, or which ions will oxidize a
particular metal. Example A solution contains
cobalt II ions (Co2). Which of the following
metals would be expected to react if placed in
the solution? Cu, Fe, Ni, Zn
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A solution contains cobalt II ions (Co2). Which
of the following metals would be expected to
react if placed in the solution? Cu, Fe, Ni,
Zn Metals above cobalt in the activity series
have a stronger tendency to undergo oxidation and
form ions. Those below have a weaker tendency to
form ions than cobalt. So - Will react - Fe,
Zn Co2(aq) Fe(s) ? Co(s) Fe2(aq)
Co2(aq) Zn(s) ? Co(s) Zn2(aq) Will not
react - Cu, Ni
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Hydrogen Ion A solution containing hydrogen ion
(H) will, according to the activity series,
oxidize any metal above it (so Pb, Sn, Ni, )
Sn(s) 2 H(aq) ? Sn2(aq) H2(g)
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Balancing Simple Oxidation-Reduction
Reactions The following three step procedure
can be used to balance simple oxidation-reduction
reactions. A variation of this method will be
used later to balance more complicated
oxidation-reduction reactions. 1) Find the
oxidation numbers for all reactants and
products 2) Write a half-reaction for oxidation,
and a half-reaction for reduction. Use electrons
for charge balance. a) Oxidation - will produce
electrons b) Reduction - will consume
electrons 3) Combine the two half reactions to
get the net ionic equation. This may mean
multiplying one or both of the half reactions to
get the electrons to cancel.
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Example Balance the following oxidation-reduction
reactions. a) Zn(s) Cu2(aq) ? Zn2(aq)
Cu(s) b) Cr(s) Pb2(aq) ? Cr3(aq) Pb(s)
55
a) Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
0 2
2 0 oxidation Zn(s) ?
Zn2(aq) 2 e- reduction Cu2(aq) 2 e- ?
Cu(s) NET Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
b) Cr(s) Pb2(aq) ? Cr3(aq) Pb(s)
0 2
3 0 oxidation (Cr(s)
? Cr3(aq) 3 e- ) x 2 reduction (Pb2(aq) 2
e- ? Pb(s) ) x 3 NET 2 Cr(s) 3 Pb2(aq) ?
2 Cr3(aq) 3 Pb(s)
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Classifying Redox Reactions Redox reactions can
be classified into various categories. Combinatio
n reaction - Two or more reactants form a simple
product Example H2(g) Cl2(g) ? 2
HCl(g) Decomposition reaction - One reactant
forms two or more products Example 2 KClO3(s)
? 2KCl(s) 3 O2(g) Disproportionation reaction
- A decomposition reaction where the same
substance is both oxidized and reduced Example
2 H2O2(aq) ? 2 H2O(l) O2(g)
57
Classifying Redox Reactions (continued) Combustio
n reaction - A single substance reacts with
oxygen to form combustion products Example CH4(g
) 2 O2(g) ? CO2(g) 2 H2O(l) In combustion
reactions molecular oxygen (O2) is reduced from
an oxidation number of 0 to an oxidation number
of -2. Displacement reaction - A reaction where
one species replaces another species
Example Zn(s) CuCl2(aq) ? ZnCl2(aq) 3
Cu(s) Br2(aq) 2 KI(aq) ? 2 KBr(aq)
I2(aq)
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Molarity The most common unit of concentration
for laboratory work is molarity. By definition,
the molarity of a solution (M) is M moles
solute liters of solution Unfortunately,
we use the same symbol, M, for molarity and for
molecular mass. The meaning of M must therefore
be determined from the context. To avoid
confusion, we often use square brackets to
indicate molarity (for example, KNO3 for the
concentration of a potassium nitrate solution, in
moles/L. Example Find the molarity of NaCl in
a solution prepared by dissolving 5.000 g NaCl
(M 58.44 g/mol) in water to form a solution
with final volume V 100.0 mL.
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Example Find the molarity of NaCl in a solution
prepared by dissolving 5.000 g NaCl (M 58.44
g/mol) in water to form a solution with final
volume V 100.0 mL. moles NaCl 5.000 g 1
mol 0.08556 mol
58.44 g liters of
solution 100.0 mL 1 L 0.1000 L

1000.0 mL So NaCl 0.08556 mol
NaCl 0.8556 mol/L 0.8556 M NaCl
0.1000 L soln
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Solution Preparation Solutions of known
concentration can easily be prepared using an
analytical balance and standard laboratory
glassware.
61
Stock Solutions and Dilutions A stock solution
is a solution of high concentration used to
prepare more dilute solution by combining known
amounts of the stock solution with pure solvent.
If the concentration of the stock solution is
known and the method used to carry out the
preparation of the dilute solution is also known,
the concentration of the dilute solution may be
found. Example A solution is prepared by
mixing 20.00 mL of the 0.8556 M NaCl solution
discussed above with water to form a dilute
solution with a final volume equal to 500.0 mL.
What is the concen-tration of this dilute
solution?
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A solution is prepared by mixing 20.00 mL of
the 0.8556 M NaCl solution discussed above with
water to form a dilute solution with a final
volume equal to 500.0 mL. What is the
concentration of this dilute solution? moles
NaCl 0.02000 L 0.8556 mol 0.01711 mol
NaCl
L stock soln M dilute soln 0.01711
mol NaCl 0.03422 mol/L
0.5000L
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General Relationship For Dilution of Stock
Solutions Based on the method used in the above
calculation a general relationship useful in
preparing dilutions of stock solutions may be
found. Let Mc molarity of initial
(concentrated) solution Lc volume, in liters,
of initial (concentrated) solution Md
molarity of final (dilute) solution Ld
volume, in liters, of final (dilute) solution
Then Mc Lc Md Ld This relationship is
based on conservation of mass, in this case, the
mass (or number of moles) of solute in the
solutions. It will work for any volume unit, as
long as the same unit is used for Lc and Ld.
64
We can check this relationship using our
previous example. A solution is prepared by
mixing 20.00 mL of the 0.8556 M NaCl solution
discussed above with water to form a dilute
solution with a final volume equal to 500.0 mL.
What is the concentration of this dilute
solution? We previously found M 0.03422
mol/L. Since Mc Lc Md Ld, then if we divide
both sides by Ld, we get Md Mc Lc (0.8556
M) (20.00 mL) 0.03422 mol/L
Ld (500.0 mL) the same as
before.
65
One of the main uses of this relationship is to
use stock solutions of known concentration and
volumetric glassware (pipettes, flasks) to
prepare more dilute solutions with particular
concentrations. Example A student has a 0.4846
M aqueous stock solution of glucose, and wishes
to prepare 100.0 mL of a 0.1000 M glucose
solution. How can she do this?
66
A student has a 0.4846 M aqueous stock solution
of glucose, and wishes to prepare 100.0 mL of a
0.1000 M glucose solution. How can she do
this? We could use the method used earlier, but
is is faster and more simple to use the
relationship Mc Lc Md Ld Mc 0.4846 M Md
0.1000 M Lc ? Ld 100.0 mL So (0.4846
M) Lc (0.1000 M) (100.0 mL) Lc (0.1000 M)
(100.0 mL) 20.6 mL
(0.4846 M) So add 20.6 mL of the stock solution
to a 100.0 mL volumetric flask, and then add
water to obtain the final volume of solution.
67
Serial Dilution It is difficult to prepare very
dilute solutions (concentration lt 10-3 M)
directly, as it would involve precisely weighing
out a small mass of solute. Such dilute
solutions are prepared by the process of
successive, or serial, dilution of a stock
solution. Example Starting with a 0.1000 M
stock solution of KNO3, prepare a dilute solution
with a concentration of 1.00 x 10-5 M.
68
Example Starting with a 0.1000 M stock
solution of KNO3, prepare a dilute solution with
a concentration of 1.00 x 10-5 M. Stock
solution 0.1000 M Take 5.00 mL of stock solution
and dilute to a final volume of 500.0 mL. Md
Mc Lc (0.1000 M) (5.00 mL) 0.001000 M
Ld
(500.0 mL) Dilute this new solution in the same
way as in the previous step Md Mc Lc
(0.00100 M) (5.00 mL) 1.00 x 10-5 M
Ld (500.0 mL)
69
Solution Stoichiometry If we know the
concentration of a solution of a strong
electrolyte, we can use the stoichiometry of the
dissolution reaction to find the concentration of
individual ions in solution. Example. A student
has a 0.0200 M solution of copper II nitrate
(Cu(NO3)2 ). What are the concentrations of Cu2
ion and NO3- ion in the solution? Dissolution
equation is Cu(NO3)2(s) ? Cu2(aq) 2
NO3-(aq) Cu2 0.0200 M Cu(NO3)2 1 mol
Cu2 0.0200 M Cu2 ion
1 mol Cu(NO3)2
NO3- 0.0200 M Cu(NO3)2 2 mol NO3-
0.0400 M NO3- ion
1 mol Cu(NO3)2
70
Gravimetric Analysis In gravimetric analysis, we
use mass measurements to determine composition or
concentrations for unknowns. One common
gravimetric method is to add a compound to a
solution to form a precipitate, which can then be
filtered out to determine the mass of solid
formed. Example An aqueous solution contains
an unknown concentra-tion of silver nitrate
(AgNO3). Excess solid sodium chloride (NaCl) is
added to a 200.00 mL sample of the silver nitrate
solution, and 0.184 g of precipitate forms. What
is the concentration of silver nitrate in the
original solution?
71
Example An aqueous solution contains an unknown
concentra-tion of silver nitrate (AgNO3). Excess
solid sodium chloride (NaCl) is added to a 200.00
mL sample of the silver nitrate solution, and
0.184 g of precipitate forms. What is the
concentration of silver nitrate in the original
solution? AgNO3(aq) ? Ag(aq) NO3-(aq)
AgNO3(aq) NaCl(aq) ? AgCl(s)
NaNO3(aq) M(AgCl) 143.32 g/mol mol AgNO3
0.184 g AgCl 1 mol AgCl 1 mol AgNO3
143.32 g
1 mol AgCl 0.00128 mol AgNO3
AgNO3 0.00128 0.00642 M
0.2000L
72
Titration The term titration refers to a
reaction between two substances where one
reactant is slowly added to a fixed amount of a
second reactant until complete reaction
occurs. Titrations are often used to find the
concentrations of solutions.
In an acid-base titration, for example, base is
slowly added to a fixed amount of acid until
complete reaction occurs (the titration can also
be carried out by the slow addition of acid to a
fixed amount of base). An indicator (a molecule
that has a different color in acid and base
solution) is used to visually estimate the
equivalence point of the titration (the point
where the reaction is complete). This should
occur as close as possible to the end point of
the titration, the point where the indicator
color change occurs.
73
Calculations Involving Titrations Since
titrations are chemical reactions, we may carry
out calculations associated with titrations in
the same way as other calculations involving
chemical reactions. We require a balanced
chemical equation and information about the
amounts of the acid and base solutions used in
the titration and the concentration of one of the
solutions (or the number of moles of one of the
reactants). Example A 20.00 mL sample of an HCl
solution of unknown concentration is titrated
with a 0.3145 M NaOH solution. The equivalence
point for the titration occurs after the addition
of 28.33 mL of the NaOH solution. What is the
concentration of the HCl solution (in units of
mol/L)? HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
74
A 20.00 mL sample of an HCl solution of unknown
concentration is titrated with a 0.3145 M NaOH
solution. The equivalence point for the
titration occurs after the addition of 28.33 mL
of the NaOH solution. What is the concentration
of the HCl solution (in units of mol/L)? HCl(aq)
NaOH(aq) ? NaCl(aq) H2O(l) Strategy 1)
Find the number of moles of NaOH (the solution
where you know both the volume and
concentration). 2) Use the balanced chemical
equation to find the number of moles of HCl. 3)
Find the concentration of the HCl solution, using
the number of moles (found in step 2) and the
volume of HCl solution (given in the problem).
75
A 20.00 mL sample of an HCl solution of unknown
concentration is titrated with a 0.3145 M NaOH
solution. The equivalence point for the
titration occurs after the addition of 28.33 mL
of the NaOH solution. What is the concentration
of the HCl solution (in units of
mol/L)? HCl(aq) NaOH(aq) ? NaCl(aq)
H2O(?) n(NaOH) 0.02833 L 0.3145 mol
8.910 x 10-3 mol NaOH
L soln n(HCl) 8.910 x 10-3
mol NaOH 1 mol HCl 8.910 x 10-3 mol HCl

1 mol NaOH M(HCl) 8.910 x 10-3 mol HCl
0.4455 mol/L 0.4455 M
0.02000 L soln
76
Oxidation-Reduction (Redox) Titration An
oxidation-reduction titration (redox titration)
is a titration based on an oxidation-reduction
reaction. Calculations are done in the same way
as in acid-base titrations. Example A 20.00 mL
sample of a solution containing Fe2 ion is
titrated with a 0.02418 M solution of BrO3- ion.
The equivalence point of the titration occurs
after the addition of 38.67 mL of the BrO3-
solution. What is the concentration of the Fe2
solution? 6 Fe2(aq) BrO3-(aq) 6 H(aq) ?
6 Fe3(aq) Br-(aq) 3 H2O(?)
77
Example A 20.00 mL sample of a solution
containing Fe2 ion is titrated with a 0.02418 M
solution of BrO3- ion. The equivalence point of
the titration occurs after the addition of 38.67
mL of the BrO3- solution. What is the
concentration of the Fe2 solution? 6 Fe2(aq)
BrO3-(aq) 6 H(aq) ? 6 Fe3(aq) Br-(aq) 3
H2O(?) Strategy 1) Find the number of moles of
BrO3- (the solution where you know both the
volume and concentration). 2) Use the balanced
chemical equation to find the number of moles of
Fe2. 3) Find the concentration of the Fe2
solution, using the number of moles (found in
step 2) and the volume of Fe2 solution (given in
the problem).
78
6 Fe2(aq) BrO3-(aq) 6 H(aq) ? 6
Fe3(aq) Br-(aq) 3 H2O(?) 20.00 mL Fe2
soln 38.67 mL BrO3- soln 0.02418 M BrO3-
moles BrO3- 0.03867 L soln 0.02418 mol
9.350 x 10-4 mol BrO3-
L soln moles Fe2
9.350 x 10-4 mol BrO3- 6 mol Fe2 5.610 x
10-3 mol Fe2
1 mol BrO3- molarity of
Fe2 soln 5.610 x 10-3 mol Fe2 0.2805
M
0.02000L soln
79
End of Chapter 4 A physicist, a biologist and a
chemist went to the ocean for the first time.
The physicist saw the ocean and was fascinated by
the waves. He said he wanted to do some research
on the fluid dynamics of the waves and walked
into the ocean. He was drowned and never
returned. The biologist saw the ocean and said
that he wanted to do research on the marine flora
and fauna he saw there. He walked into the ocean
and also never returned. The chemist waited for
a long time and then wrote down the following
observation. Physicists and biologists are
soluble in ocean water. - anonymous Two
questions to distinguish chemists from
non-chemists (1) How do you pronounce
unionized? (2) What is a mole?